Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r - 1}{-2r^2 + 4r - 2} \div \dfrac{r - 5}{r^2 - 9r + 20} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{r - 1}{-2r^2 + 4r - 2} \times \dfrac{r^2 - 9r + 20}{r - 5} $ First factor out any common factors. $q = \dfrac{r - 1}{-2(r^2 - 2r + 1)} \times \dfrac{r^2 - 9r + 20}{r - 5} $ Then factor the quadratic expressions. $q = \dfrac {r - 1} {-2(r - 1)(r - 1)} \times \dfrac {(r - 5)(r - 4)} {r - 5} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(r - 1) \times (r - 5)(r - 4) } { -2(r - 1)(r - 1) \times (r - 5)} $ $q = \dfrac {(r - 5)(r - 4)(r - 1)} {-2(r - 1)(r - 1)(r - 5)} $ Notice that $(r - 1)$ and $(r - 5)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {(r - 5)(r - 4)\cancel{(r - 1)}} {-2\cancel{(r - 1)}(r - 1)(r - 5)} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $q = \dfrac {\cancel{(r - 5)}(r - 4)\cancel{(r - 1)}} {-2\cancel{(r - 1)}(r - 1)\cancel{(r - 5)}} $ We are dividing by $r - 5$ , so $r - 5 \neq 0$ Therefore, $r \neq 5$ $q = \dfrac {r - 4} {-2(r - 1)} $ $ q = \dfrac{-(r - 4)}{2(r - 1)}; r \neq 1; r \neq 5 $